Algebra Examples

p (x) = (x + 2) (2 x^{2} + 3 x - 9)

$p (x) = (x + 2) (2 x^{2} + 3 x - 9)$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in

0

$0$ for

y

$y$ and solve for

x

$x$ .

0 = (x + 2) (2 x^{2} + 3 x - 9)

$0 = (x + 2) (2 x^{2} + 3 x - 9)$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as

(x + 2) (2 x^{2} + 3 x - 9) = 0

$(x + 2) (2 x^{2} + 3 x - 9) = 0$ .

(x + 2) (2 x^{2} + 3 x - 9) = 0

$(x + 2) (2 x^{2} + 3 x - 9) = 0$

Step 1.2.2

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x + 2 = 0

$x + 2 = 0$

2 x^{2} + 3 x - 9 = 0

$2 x^{2} + 3 x - 9 = 0$

Step 1.2.3

Set

x + 2

$x + 2$ equal to

0

$0$ and solve for

x

$x$ .

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Step 1.2.3.1

Set

x + 2

$x + 2$ equal to

0

$0$ .

x + 2 = 0

$x + 2 = 0$

Step 1.2.3.2

Subtract

2

$2$ from both sides of the equation.

x = - 2

$x = - 2$

x = - 2

$x = - 2$

Step 1.2.4

Set

2 x^{2} + 3 x - 9

$2 x^{2} + 3 x - 9$ equal to

0

$0$ and solve for

x

$x$ .

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Step 1.2.4.1

Set

2 x^{2} + 3 x - 9

$2 x^{2} + 3 x - 9$ equal to

0

$0$ .

2 x^{2} + 3 x - 9 = 0

$2 x^{2} + 3 x - 9 = 0$

Step 1.2.4.2

Solve

2 x^{2} + 3 x - 9 = 0

$2 x^{2} + 3 x - 9 = 0$ for

x

$x$ .

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Step 1.2.4.2.1

Factor by grouping.

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Step 1.2.4.2.1.1

For a polynomial of the form

a x^{2} + b x + c

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

a \cdot c = 2 \cdot - 9 = - 18

$a \cdot c = 2 \cdot - 9 = - 18$ and whose sum is

b = 3

$b = 3$ .

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Step 1.2.4.2.1.1.1

Factor

3

$3$ out of

3 x

$3 x$ .

2 x^{2} + 3 (x) - 9 = 0

$2 x^{2} + 3 (x) - 9 = 0$

Step 1.2.4.2.1.1.2

Rewrite

3

$3$ as

- 3

$- 3$ plus

6

$6$

2 x^{2} + (- 3 + 6) x - 9 = 0

$2 x^{2} + (- 3 + 6) x - 9 = 0$

Step 1.2.4.2.1.1.3

Apply the distributive property.

2 x^{2} - 3 x + 6 x - 9 = 0

$2 x^{2} - 3 x + 6 x - 9 = 0$

2 x^{2} - 3 x + 6 x - 9 = 0

$2 x^{2} - 3 x + 6 x - 9 = 0$

Step 1.2.4.2.1.2

Factor out the greatest common factor from each group.

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Step 1.2.4.2.1.2.1

Group the first two terms and the last two terms.

(2 x^{2} - 3 x) + 6 x - 9 = 0

$(2 x^{2} - 3 x) + 6 x - 9 = 0$

Step 1.2.4.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

x (2 x - 3) + 3 (2 x - 3) = 0

$x (2 x - 3) + 3 (2 x - 3) = 0$

x (2 x - 3) + 3 (2 x - 3) = 0

$x (2 x - 3) + 3 (2 x - 3) = 0$

Step 1.2.4.2.1.3

Factor the polynomial by factoring out the greatest common factor,

2 x - 3

$2 x - 3$ .

(2 x - 3) (x + 3) = 0

$(2 x - 3) (x + 3) = 0$

(2 x - 3) (x + 3) = 0

$(2 x - 3) (x + 3) = 0$

Step 1.2.4.2.2

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

2 x - 3 = 0

$2 x - 3 = 0$

x + 3 = 0

$x + 3 = 0$

Step 1.2.4.2.3

Set

2 x - 3

$2 x - 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 1.2.4.2.3.1

Set

2 x - 3

$2 x - 3$ equal to

0

$0$ .

2 x - 3 = 0

$2 x - 3 = 0$

Step 1.2.4.2.3.2

Solve

2 x - 3 = 0

$2 x - 3 = 0$ for

x

$x$ .

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Step 1.2.4.2.3.2.1

Add

3

$3$ to both sides of the equation.

2 x = 3

$2 x = 3$

Step 1.2.4.2.3.2.2

Divide each term in

2 x = 3

$2 x = 3$ by

2

$2$ and simplify.

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Step 1.2.4.2.3.2.2.1

Divide each term in

2 x = 3

$2 x = 3$ by

2

$2$ .

\frac{2 x}{2} = \frac{3}{2}

$\frac{2 x}{2} = \frac{3}{2}$

Step 1.2.4.2.3.2.2.2

Simplify the left side.

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Step 1.2.4.2.3.2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 1.2.4.2.3.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Step 1.2.4.2.3.2.2.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{3}{2}$

Step 1.2.4.2.4

Set

$x + 3$ equal to

$0$ and solve for

$x$ .

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Step 1.2.4.2.4.1

Set

$x + 3$ equal to

$0$ .

$x + 3 = 0$

Step 1.2.4.2.4.2

Subtract

$3$ from both sides of the equation.

$x = - 3$

Step 1.2.4.2.5

The final solution is all the values that make

$(2 x - 3) (x + 3) = 0$ true.

$x = \frac{3}{2}, - 3$

Step 1.2.5

The final solution is all the values that make

$(x + 2) (2 x^{2} + 3 x - 9) = 0$ true.

$x = - 2, \frac{3}{2}, - 3$

Step 1.3

x-intercept(s) in point form.

x-intercept(s):

$(- 2, 0), (\frac{3}{2}, 0), (- 3, 0)$

x-intercept(s):

$(- 2, 0), (\frac{3}{2}, 0), (- 3, 0)$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in

$0$ for

$x$ and solve for

$y$ .

$y = ((0) + 2) (2 {(0)}^{2} + 3 (0) - 9)$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = (0 + 2) (2 {(0)}^{2} + 3 (0) - 9)$

Step 2.2.2

Remove parentheses.

$y = (0 + 2) (2 \cdot 0^{2} + 3 (0) - 9)$

Step 2.2.3

Remove parentheses.

$y = ((0) + 2) (2 {(0)}^{2} + 3 (0) - 9)$

Step 2.2.4

Simplify

$((0) + 2) (2 {(0)}^{2} + 3 (0) - 9)$ .

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Step 2.2.4.1

Add

$0$ and

$2$ .

$y = 2 (2 {(0)}^{2} + 3 (0) - 9)$

Step 2.2.4.2

Simplify each term.

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Step 2.2.4.2.1

Raising

$0$ to any positive power yields

$0$ .

$y = 2 (2 \cdot 0 + 3 (0) - 9)$

Step 2.2.4.2.2

Multiply

$2$ by

$0$ .

$y = 2 (0 + 3 (0) - 9)$

Step 2.2.4.2.3

Multiply

$3$ by

$0$ .

$y = 2 (0 + 0 - 9)$

Step 2.2.4.3

Simplify the expression.

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Step 2.2.4.3.1

Add

$0$ and

$0$ .

$y = 2 (0 - 9)$

Step 2.2.4.3.2

Subtract

$9$ from

$0$ .

$y = 2 \cdot - 9$

Step 2.2.4.3.3

Multiply

$2$ by

$- 9$ .

$y = - 18$

Step 2.3

y-intercept(s) in point form.

y-intercept(s):

$(0, - 18)$

y-intercept(s):

$(0, - 18)$

Step 3

List the intersections.

x-intercept(s):

$(- 2, 0), (\frac{3}{2}, 0), (- 3, 0)$

y-intercept(s):

$(0, - 18)$

Step 4